Factor completely. $(x^2-4)(x^2+6x+9)=$
The expression is already somewhat factored, because it is given as the product of two factors, $(x^2-4)$ and $(x^2+6x+9)$. To completely factor the expression, we need to factor each of these factors further. Factoring $(x^2-4)$ We notice this expression has the difference of squares pattern: $\begin{aligned} &\phantom{=}x^2-4 \\\\ &=(x)^2-(2)^2 \\\\ &=(x+2)(x-2) \end{aligned}$ Factoring $(x^2+6x+9)$ We notice this expression has the perfect square pattern: $\begin{aligned} &\phantom{=}x^2+6x+9 \\\\ &=(x)^2+2(x)(3)+(3)^2 \\\\ &=(x+3)^2 \end{aligned}$ Putting it all together $\begin{aligned} &\phantom{=}{(x^2-4)}C{(x^2+6x+9)} \\\\ &={(x+2)(x-2)}C{(x+3)^2} \end{aligned}$ In conclusion, this is the completely factored expression: $(x+2)(x-2)(x+3)^2$